Let z be a complex number satisfying $$\arg\bigg(\frac{z^3-1}{z^3+1}\bigg) = \frac{\pi}{2}$$ on the Argand plane. Then, length of the arc of the locus of z for which $Re(z)>0$ and $Img(z)<0$ (where '$Re$' and '$Img$' represent real and imaginary part respectively) is?
Since it's argument is $\frac{\pi}{2}$, the complex number will be purely imaginary.
So, its conjugate would be equal to the negative of itself.
Solving that gives me, $|z|^6 = 1$
Since $|z|$ is a positive real number, so the only solution is $|z| = 1$
So, the answer should be $\frac{\pi}{2}$
But the answer given is $\frac{\pi}{6}$
Any help would be appreciated.
We have \begin{align}\operatorname{arg}\left(\frac{z^3-1}{z^3+1}\right)=\frac{\pi}{2}&\iff \exists \lambda >0,\frac{z^3-1}{z^3+1}=\lambda i \\&\iff \exists \lambda>0, z^3=\frac{1+\lambda i}{1-\lambda i} \\&\iff \exists \lambda>0, z^3=\frac{(1+\lambda i)^2}{1+\lambda^2} \\&\iff \exists \lambda>0, z^3=\frac{(1-\lambda^2)+2\lambda i}{1+\lambda^2} \\&\iff |z|=1 \wedge \Im z^3>0 \\&\iff |z|=1 \wedge \operatorname{arg} z^3\in (0,\pi) \\&\iff \exists \theta \in(0,\pi/3)\cup (2\pi/3,\pi)\cup (4\pi/3,5\pi/3),z=e^{i\theta}.\end{align} If in addition, $\Re z>0$ and $\Im z<0$, then $$\operatorname{arg}\left(\frac{z^3-1}{z^3+1}\right)=\frac{\pi}{2}\iff \exists \theta\in (3\pi/2,5\pi/3), z=e^{i\theta}.$$ So, your book is right, the answer is $5\pi/3-3\pi/2=\pi/6$.